Question

$PQRS$ is the smallest square whose vertices are on the respective sides of the square $ABCD$. The ratio of the areas of $\square PQRS$ to $\square ABCD$ is

• A. $1:2$
• B. $1:\sqrt{2}$
• C. $1:3$
• D. $2:3$

Answer 1

Answer: (A)

$1:2$

To forma square inside the square, take a midpoint on every sides of square and join them to form a square.

As $PQRS$ is smallest square of $\square ABCD$.

So, $P,Q,R$ and $S$ will be midpoints of $AB,BC,CD$ and $DA$ respectively.

$AB=2AP,BC=2BQ,CD=2CR,DA=2DS$

Suppose sides of $\square ABCD$ is $x$.

$\therefore Area\left(\square ABCD\right)=x^2$

$P{S}^2=A{P}^2+A{S}^2=(x/2{)}^2+(x/2{)}^2=\frac{x^2}{2}$

$\therefore PS=\frac{x}{\sqrt{2}}$

$\therefore$ $Area\left(\square PQRS\right)=sid{e}^2={\left(\frac{x}{\sqrt{2}}\right)}^2=\frac{x^2}{2}$.

$\therefore \frac{A\left(\square PQRS\right)}{A\left(\square ABCD\right)}=\frac{\frac{x^2}{2}}{x^2}=\frac{1}{2}$.

$A\left(\square PQRS\right):A\left(\square ABCD\right)=1:2$.

Hence, option A is correct.